Question

диференциальное исчисление

Answer 1

Ответ:

1.

$y = 5 {x}^{2} - {x}^{ \frac{4}{3} } + 4 {x}^{ - 3} - 5 {x}^{ - 1}$

$y' = 10x - \frac{4}{3} {x}^{ \frac{1}{3} } - 12 {x}^{ - 4 } + 5 {x}^{ - 2} = \\ = 10x - \frac{4}{3} \sqrt[3]{x} - \frac{12}{ {x}^{4} } + \frac{5}{ {x}^{2} }$

2.

$y = \frac{ \sin(t) }{1 + \cos(t) }$

$y' = \frac{ \cos(t)(1 + \cos(t)) - ( - \sin(t) ) \times \sin(t) }{ {(1 + \cos(t)) }^{2} } \\ = \frac{ \cos(t) + { \cos }^{2} (t) + { \sin }^{2} (t)}{ {(1 + \cos(t)) }^{2} } = \\ = \frac{ \cos(t) }{ {(1 + \cos(t)) }^{2} }$

3.

$y = x \sin(x) ln(x)$

$y' = (x \sin(x) )' ln(x) + ( ln(x))' x \sin(x) = \\ = (x' \sin(x) + ( \sin(x) )'x) ln(x) + \frac{1}{x} \times x \sin(x) = \\ = ( \sin(x) + x \cos(x) ) \times ln(x) + \sin(x)$

4.

$y = {arccos}^{2} (4x) \times ln(x - 3)$

$y' = 2arccos(4x) \times ( - \frac{1}{ \sqrt{ 1 - 16 {x}^{2} } } ) ln(x - 3) + \frac{1}{x - 3} + {arccos}^{2} (4x) = \\ = arccos(4x) \times ( - \frac{2 ln(x - 3) }{ \sqrt{1 - 16 {x}^{2} } } + \frac{arccos(4x)}{x - 3} )$